Rigid Rotor

The rigid rotor is the pure angular-motion problem. In the simplest linear rotor, the radius or bond length is fixed, so the only remaining degrees of freedom are the orientation angles on a sphere. The Hamiltonian is proportional to the angular-momentum Casimir:

H_{\rm rot} = \frac{L^2}{2I}.

This makes the rotor a clean model of how rotational symmetry organizes quantum states. The full wave-mechanics model, normalization, molecular interpretation, and spectroscopy details live in Rigid Rotor in the Wave Mechanics volume. This page focuses on the symmetry content.

Canonical Split

The rigid rotor sits at the intersection of model physics and angular-momentum representation theory.

Question	Canonical home
What is the full rotor model and its molecular interpretation?	Rigid Rotor
Why are the eigenstates spherical harmonics?	Spherical Harmonics
Why does the energy depend on $J(J+1)$ ?	this page
How do real rotational spectra and external fields modify the ideal model?	Rotational Spectra and Rotor in External Fields

Here “rigid rotor” means the ideal linear rotor with fixed moment of inertia $I$ . More general symmetric and asymmetric tops have richer rotational Hamiltonians and are not treated as the main model here.

Configuration Space

A linear rotor orientation is a point on the sphere:

\Omega=(\theta,\phi)\in S^2.

The Hilbert space is

L^2(S^2,d\Omega),

with

d\Omega=\sin\theta\,d\theta\,d\phi.

The wavefunction $\Psi(\theta,\phi)$ is a scalar function on the sphere. For this ideal scalar rotor, ordinary single-valuedness on $S^2$ implies integer angular-momentum labels. Spinor rotor states, nuclear exchange restrictions, and molecular symmetry constraints are separate refinements.

Hamiltonian as a Casimir

The rotational kinetic energy of an isotropic linear rotor is

H_{\rm rot} = \frac{L^2}{2I},

where $I$ is the moment of inertia. Since

L^2=-\hbar^2\Delta_{S^2},

the Hamiltonian is the angular Laplacian on the sphere multiplied by $-\hbar^2/(2I)$ .

The important symmetry point is that $L^2$ is the Casimir operator of the rotation algebra. It commutes with every component:

[L^2,L_i]=0.

Therefore

[H_{\rm rot},L_i]=0, \qquad i=x,y,z.

The ideal rotor is fully rotationally invariant. Its energy cannot depend on the orientation label $M$ relative to an arbitrary chosen $z$ axis.

Eigenstates and Labels

The simultaneous eigenstates of $L^2$ and $L_z$ are spherical harmonics:

\Psi_{JM}(\theta,\phi) = Y_J^M(\theta,\phi).

They satisfy

L^2Y_J^M = \hbar^2J(J+1)Y_J^M,

and

L_zY_J^M = \hbar M Y_J^M.

The allowed labels for an ordinary scalar rotor are

J=0,1,2,\ldots,

and

M=-J,-J+1,\ldots,J.

This page uses the molecular-rotor convention $J,M$ . The same mathematical labels are often written $\ell,m$ in central-potential wave mechanics.

Energy Levels

Acting with $H_{\rm rot}$ on $Y_J^M$ gives

H_{\rm rot}Y_J^M = \frac{\hbar^2}{2I}J(J+1)Y_J^M.

Thus

E_J = \frac{\hbar^2}{2I}J(J+1).

It is common to define the rotational constant

B=\frac{\hbar^2}{2I},

so that

E_J=BJ(J+1).

The spacing between adjacent levels is

E_{J+1}-E_J = 2B(J+1).

The levels are not evenly spaced in energy. The transition lines of the ideal dipole rotor can still form a regular ladder because allowed transitions connect adjacent $J$ values.

Degeneracy

For fixed $J$ , there are

2J+1

allowed $M$ values. Since $H_{\rm rot}$ depends only on $L^2$ , all $M$ states in the same $J$ multiplet have the same energy:

g_J=2J+1.

This degeneracy is the representation-theoretic signature of rotational invariance. A weak external electric or magnetic field selects a physical axis and can split the $M$ levels. Anisotropic molecular environments or nonideal rotor terms can also break the simple degeneracy.

Relation to Central Potentials

The angular part of a central-potential wavefunction is also a spherical harmonic:

\psi(r,\theta,\phi) = R_{\alpha\ell}(r)Y_\ell^m(\theta,\phi).

The difference is that a central-potential particle still has radial motion. The rigid rotor freezes the radial coordinate, so the angular factor is the whole configuration-space wavefunction.

This makes the rotor useful pedagogically:

it isolates the $L^2$ eigenvalue problem;
it shows the $2J+1$ degeneracy without a radial equation;
it gives a physical model where spherical harmonics are literal stationary states;
it prepares the selection-rule machinery used in rotational spectroscopy.

Selection-Rule Preview

For a polar linear rotor, electric-dipole rotational transitions are controlled by how the dipole operator transforms under rotations. The dipole is a vector operator, so it carries angular momentum rank one.

The basic rotational selection rule for the ideal linear rotor is

\Delta J=\pm1,

with $\Delta M=0,\pm1$ depending on polarization and the chosen quantization axis. The detailed tensor-operator derivation belongs to Applications to Molecular Rotations and the line-position application belongs to Rotational Spectra.

This rule is not a statement that the rotor can only have neighboring energy levels. It is a statement about which matrix elements of the dipole operator are allowed by angular momentum.

Common Mistakes

Using half-integer $J$ labels for an ordinary scalar rotor on $S^2$ .
Forgetting that the $M$ degeneracy follows from rotational invariance and can be split by external fields.
Confusing the rotational constant $B$ with a magnetic field.
Treating the spherical harmonic shape as a classical orientation of a rigid rod.
Assuming the ideal rotor includes vibration, centrifugal distortion, spin, nuclear exchange symmetry, or electronic structure.
Applying $\Delta J=\pm1$ as a universal rule for every perturbation; it is the electric-dipole rule for the ideal linear polar rotor.

Cross-Links

References

A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
R. N. Zare, Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics, Wiley, 1988.
J. M. Brown and A. Carrington, Rotational Spectroscopy of Diatomic Molecules, Cambridge University Press, 2003.
R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

Find the energy and degeneracy of the $J=2$ level.

Solution

The energy formula is

E_J=BJ(J+1).

For $J=2$ ,

E_2=B(2)(3)=6B.

The degeneracy is

g_J=2J+1,

g_2=5.

The allowed $M$ values are

M=-2,-1,0,1,2.

Explain why the ideal rotor energy does not depend on $M$ .

Solution

The ideal rotor Hamiltonian is

H_{\rm rot}=\frac{L^2}{2I}.

It depends on the Casimir $L^2$ , not on any chosen component such as $L_z$ . Since

L^2Y_J^M=\hbar^2J(J+1)Y_J^M,

all states with the same $J$ and different $M$ have the same energy. Physically, the free rotor has no preferred spatial axis, so different orientations of the angular momentum multiplet are degenerate.

Compare the rigid rotor with a central-potential bound state.

Solution

Both use spherical harmonics as angular eigenfunctions. For a central potential,

\psi_{\alpha\ell m}(r,\theta,\phi) = R_{\alpha\ell}(r)Y_\ell^m(\theta,\phi),

so the radial function carries additional dynamics. For the rigid rotor, the radius is fixed and the wavefunction is just

\Psi_{JM}(\theta,\phi)=Y_J^M(\theta,\phi).

Thus the rotor isolates the angular part of the central-potential separation problem.

Why does the ideal scalar rotor have integer $J$ values rather than half-integer values?

Solution

The ideal scalar rotor wavefunction is an ordinary single-valued function on $S^2$ . Its angular eigenfunctions are spherical harmonics, which carry integer angular momentum labels:

J=0,1,2,\ldots.

Half-integer labels arise from spinor representations of the $SU(2)$ double cover, not from ordinary scalar functions on the sphere.